CUET UG Physics Booster Test 2- Extrinsic and Doping Concepts -

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CUET UG Physics Booster Test 2- Extrinsic and Doping Concepts

⏱️ 30 Minutes

❓ 20 Questions

📌 Answers are locked once submitted — results and explanations appear at the end.

CUET UG Physics Booster Test 2- Extrinsic and Doping Concepts

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QUESTION 1 OF 20

The reason intrinsic semiconductors require impurity additions

Answer Review

1 The reason intrinsic semiconductors require impurity additions

A depends on the size of the original lattice

B depends on the angle between covalent bonds

C is to overcome the fact that at room temperature its conductivity is very low for developing electronic devices

D independent of the intrinsic carrier concentration

💡 Short Explanation

Intrinsic semiconductors have low conductivity. Few charge carriers are available. Doping increases conductivity significantly.

✅ Correct Answer Explanation

Pure semiconductors contain only thermally generated charge carriers. Their conductivity at room temperature is too low for most practical electronic applications. Therefore, impurities are intentionally added to increase carrier concentration and improve conductivity.

❓ Why Other Options Are Incorrect

Option A → Lattice size is not the primary reason.
Option B → Bond angles do not determine the need for doping.
Option D → Intrinsic carrier concentration directly affects conductivity.

🎯 Strategy

Used

Conceptual Matching

Application:

Identify why doping is introduced.

Final Logic:

Low intrinsic conductivity necessitates impurity addition.

🧠 Memory Trick

"Low Conductivity → Add Impurities."

📚 Class XII – Physics Part II – Chapter: Semiconductor Electronics – Topic: Need for Doping

2 If a few parts per million of a suitable impurity are added to a pure semiconductor yielding an extrinsic conductivity σext compared to intrinsic σint, the correct relation will be

A σext − σint = 0

B σext >> σint

C σint >> σext

D σext + σint = 0

💡 Short Explanation

Doping greatly increases carrier concentration. Conductivity rises manifold. Extrinsic conductivity is much larger.

✅ Correct Answer Explanation

Even a very small concentration of donor or acceptor impurities contributes a large number of charge carriers. Consequently, the conductivity of the extrinsic semiconductor becomes much greater than that of the intrinsic semiconductor. \({\sigma}_{ext}≫{\sigma}_{int}\)

❓ Why Other Options Are Incorrect

Option A → Conductivities are not equal.
Option C → Reverse of the correct relation.
Option D → No physical meaning.

🎯 Strategy

Used

Conceptual Comparison

Application:

Compare intrinsic and extrinsic conductivities.

Final Logic:

Doping increases conductivity dramatically.

🧠 Memory Trick

"Doping ⇒ Conductivity Jumps."

📚 Class XII – Physics Part II – Chapter: Semiconductor Electronics – Topic: Extrinsic Conductivity

3 Dopant selection statements
1. Si and Ge belong to the fourth group in the Periodic table.
2. The dopant element is chosen from the nearby fifth or third group.
3. The size of the dopant atom must be drastically different from Si or Ge.
4. The dopant distorts the original pure semiconductor lattice substantially.

A (A), (B)

B (B), (C)

C (C), (D)

D (A), (D)

💡 Short Explanation

Si and Ge are Group IV elements. Dopants come from Group III or Group V. Dopant size should be similar.

✅ Correct Answer Explanation

Silicon and Germanium belong to Group IV of the periodic table. To produce p-type or n-type semiconductors, impurities are selected from neighboring Group III or Group V elements. Statements C and D are incorrect because the dopant should not drastically distort the crystal lattice.

❓ Why Other Options Are Incorrect

Option B → Includes incorrect statement C.
Option C → Includes incorrect statements C and D.
Option D → Includes incorrect statement D.

🎯 Strategy

Used

Option Grouping

Application:

Evaluate each statement independently.

Final Logic:

Only A and B are correct.

🧠 Memory Trick

"Group IV Host, Group III/V Dopant."

📚 Class XII – Physics Part II – Chapter: Semiconductor Electronics – Topic: Dopant Selection

4 When occupying lattice sites, a dopant with valency 5 creates a _____ semiconductor, whereas a dopant with valency 3 creates a _____ semiconductor:

A n-type, p-type

B p-type, n-type

C Direct, Indirect

D Intrinsic, Extrinsic

💡 Short Explanation

Pentavalent impurities donate electrons. Trivalent impurities create holes. They form n-type and p-type materials respectively.

✅ Correct Answer Explanation

A pentavalent impurity contributes one extra electron and forms an n-type semiconductor. A trivalent impurity creates a hole deficiency and forms a p-type semiconductor.

❓ Why Other Options Are Incorrect

Option B → Reversed order.
Option C → Unrelated classification.
Option D → Both materials are extrinsic.

🎯 Strategy

Used

Direct Recall

Application:

Recall the effect of donor and acceptor impurities.

Final Logic:

Valency 5 → n-type, Valency 3 → p-type.

🧠 Memory Trick

"5 Gives Electron, 3 Gives Hole."

📚 Class XII – Physics Part II – Chapter: Semiconductor Electronics – Topic: n-type and p-type Semiconductors

5 Incorrect statement about the size similarity constraint during doping

A The dopant must not distort the original pure semiconductor lattice.

B It must occupy only a very few of the original semiconductor atom sites.

C The sizes of the dopant and the semiconductor atoms should be nearly the same.

D Size similarity ensures that the intrinsic generation of carriers is completely halted.

💡 Short Explanation

Similar size preserves the crystal structure. Intrinsic carrier generation still occurs. Doping does not stop thermal generation.

✅ Correct Answer Explanation

The purpose of size similarity is to minimize lattice distortion when impurity atoms replace host atoms. It does not eliminate intrinsic carrier generation, which continues due to thermal excitation.

❓ Why Other Options Are Incorrect

Option A → Correct requirement.
Option B → Correct doping condition.
Option C → Correct requirement.

🎯 Strategy

Used

Elimination

Application:

Identify the statement unrelated to size similarity.

Final Logic:

Intrinsic generation continues even after doping.

🧠 Memory Trick

"Same Size ≠ No Thermal Carriers."

📚 Class XII – Physics Part II – Chapter: Semiconductor Electronics – Topic: Doping Constraints

6 If an intrinsic Si crystal with 5 × 10²⁸ atoms m⁻³ is doped at 2 ppm with a pentavalent impurity, what is the concentration of the donor atoms (ND)?

A 1 × 10²³ m⁻³

B 1 × 10²² m⁻³

C 5 × 10²² m⁻³

D 2 × 10⁶ m⁻³

💡 Short Explanation

2 ppm = 2 × 10⁻⁶. Multiply by total atom concentration. Gives donor concentration.

✅ Correct Answer Explanation

\(N_{D}=(5\times {10}^{28})(2\times {10}^{-6})N_{D}=10\times {10}^{22}N_{D}=1\times {10}^{23} m^{-3}\)

❓ Why Other Options Are Incorrect

Option B → One-tenth of correct value.
Option C → Corresponds to 1 ppm.
Option D → Much too small.

🎯 Strategy

Used

Substitution

Application:

Use ppm = parts per million.

Final Logic:

2 ppm of 5 × 10²⁸ equals 1 × 10²³ m⁻³.

🧠 Memory Trick

"ppm Means ×10⁻⁶."

📚 Class XII – Physics Part II – Chapter: Semiconductor Electronics – Topic: Dopant Concentration

7 In the formation of an n-type semiconductor through doping

A each dopant has an equal charge for certain values of capacitances only

B each dopant atom donates one extra electron for conduction and is known as a donor impurity

C each impurity has different charge for any value of its capacitance

D each donor atom acts as a minority carrier at room temperature

💡 Short Explanation

Pentavalent atoms have one extra electron. The extra electron becomes a conduction electron. Such impurities are donors.

✅ Correct Answer Explanation

Pentavalent dopants such as Phosphorus, Arsenic, and Antimony provide one extra valence electron. This electron can easily become free and contribute to conduction.

❓ Why Other Options Are Incorrect

Option A → Not relevant.
Option C → Incorrect physical statement.
Option D → Donor electrons are majority carriers.

🎯 Strategy

Used

Direct Recall

Application:

Recall donor impurity function.

Final Logic:

Donor atoms supply conduction electrons.

🧠 Memory Trick

"Donor Donates Electron."

📚 Class XII – Physics Part II – Chapter: Semiconductor Electronics – Topic: Donor Impurities

8 Correct statements about the fifth electron of a pentavalent dopant
1. Four of its electrons bond with the four silicon neighbors.
2. The fifth electron is seen as part of the effective core of the atom.
3. The ionization energy to set this electron free is about 1.1 eV for Si.
4. Even at room temperature it will be free to move in the lattice.

A (A), (B), (D)

B (A), (B), (C)

C (B), (C), (D)

D (A), (D)

💡 Short Explanation

Four electrons form bonds. Fifth electron is weakly bound. It becomes free at room temperature.

✅ Correct Answer Explanation

In a pentavalent impurity atom, four electrons participate in covalent bonding with neighboring silicon atoms. The fifth electron is only weakly bound and becomes free for conduction even at room temperature. Statement B is incorrect because the fifth electron is not part of the effective core. Statement C is incorrect because the donor ionization energy is much smaller than 1.1 eV.

❓ Why Other Options Are Incorrect

Option A → Includes incorrect statement B.
Option B → Includes incorrect statements B and C.
Option C → Includes incorrect statements B and C.

🎯 Strategy

Used

Option Grouping

Application:

Evaluate donor-electron properties.

Final Logic:

Only A and D are correct.

🧠 Memory Trick

"Four Bond, One Moves."

📚 Class XII – Physics Part II – Chapter: Semiconductor Electronics – Topic: Donor Electrons

9 Match List I with List II regarding carrier concentration sources in n-type Si

List I	List II
P. Majority electrons source	1. Only intrinsic generation
Q. Minority holes source	2. Increases intrinsic holes and electrons weakly
R. Thermal generation effect	3. Donors and intrinsic generation combined
S. Recombination effect	4. Reduces the number of holes further

A P-3, Q-1, R-2, S-4

B P-1, Q-3, R-2, S-4

C P-3, Q-1, R-4, S-2

D P-2, Q-4, R-1, S-3

💡 Short Explanation

Majority electrons come from donors and intrinsic generation. Minority holes arise intrinsically. Recombination reduces holes.

✅ Correct Answer Explanation

Correct matching: P → Donors and intrinsic generation combined (3) Q → Only intrinsic generation (1) R → Increases intrinsic holes and electrons weakly (2) S → Reduces the number of holes further (4)

❓ Why Other Options Are Incorrect

Incorrect assignment of carrier sources and effects.

🎯 Strategy

Used

Option Grouping

Application:

Match carrier-generation concepts.

Final Logic:

Only Option A correctly matches all entries.

🧠 Memory Trick

"Donors Give Electrons, Intrinsic Gives Holes."

📚 Class XII – Physics Part II – Chapter: Semiconductor Electronics – Topic: n-Type Carrier Concentration

10 The concentration of holes in an n-type semiconductor

A depends on the exact quantity of donor states

B depends on the angle between r and p

C is reduced further because the rate of recombination increases due to the abundance of electrons

D independent of separation of charges

💡 Short Explanation

n-type material contains many electrons. Recombination rate increases. Hole concentration decreases.

✅ Correct Answer Explanation

In an n-type semiconductor, donor impurities provide a large number of electrons. These electrons readily recombine with holes, thereby reducing the hole concentration below its intrinsic value.

❓ Why Other Options Are Incorrect

Option A → Not the direct reason.
Option B → Irrelevant.
Option D → Not related to carrier concentration.

🎯 Strategy

Used

Cause-and-Effect Analysis

Application:

Relate electron abundance to recombination.

Final Logic:

More electrons → more recombination → fewer holes.

🧠 Memory Trick

"Many Electrons, Few Holes."

📚 Class XII – Physics Part II – Chapter: Semiconductor Electronics – Topic: Minority Carriers in n-Type Semiconductors

11 Trivalent dopant bonding statements
1. It has one valence electron less than Si or Ge.
2. It forms covalent bonds with three neighboring Si atoms.
3. It does not have any electron to offer to the fourth Si atom.
4. It donates one extra electron to the lattice.

A (A), (B), (C)

B (A), (C), (D)

C (B), (C), (D)

D (A), (B), (D)

💡 Short Explanation

Trivalent impurities have only three valence electrons. They form three covalent bonds. One bond remains incomplete, creating a hole.

✅ Correct Answer Explanation

Trivalent impurities such as Boron and Aluminum possess three valence electrons. They form three covalent bonds with neighboring silicon atoms but cannot complete the fourth bond, resulting in a hole. Statement D is incorrect because trivalent dopants accept electrons rather than donate extra electrons.

❓ Why Other Options Are Incorrect

Option B → Includes incorrect statement D.
Option C → Includes incorrect statement D.
Option D → Includes incorrect statement D.

🎯 Strategy

Used

Option Grouping

Application:

Evaluate each statement about acceptor impurities.

Final Logic:

Only A, B, and C are correct.

🧠 Memory Trick

"Three Electrons → One Missing Bond."

📚 Class XII – Physics Part II – Chapter: Semiconductor Electronics – Topic: Trivalent Dopants

12 When doped into a tetravalent lattice, Aluminum or Boron acts as a fixed core of one _____ charge along with its associated _____:

A Positive, Hole

B Negative, Hole

C Positive, Electron

D Negative, Electron

💡 Short Explanation

Acceptor atoms gain an electron. They become negatively ionized. A hole is left behind.

✅ Correct Answer Explanation

When a trivalent impurity accepts an electron from a neighboring bond, it becomes a negatively charged acceptor ion. The missing electron in the bond behaves as a positively charged hole.

❓ Why Other Options Are Incorrect

Option A → Acceptor ion is not positive.
Option C → Associated carrier is a hole, not a free electron.
Option D → Hole is the associated carrier, not an electron.

🎯 Strategy

Used

Charge Analysis

Application:

Determine charges after electron acceptance.

Final Logic:

Acceptor ion → Negative; Carrier → Hole.

🧠 Memory Trick

"Acceptor Gains Electron → Negative Ion."

📚 Class XII – Physics Part II – Chapter: Semiconductor Electronics – Topic: Acceptor Impurities

13 Incorrect statement about majority holes in p-type materials

A One acceptor atom gives one hole.

B These holes are in addition to the intrinsically generated holes.

C Recombination process increases the number of intrinsically generated electrons (nₑ).

D The source of conduction electrons is only intrinsic generation.

💡 Short Explanation

Recombination destroys carriers. It does not increase electron concentration. Holes remain the majority carriers.

✅ Correct Answer Explanation

Recombination occurs when an electron combines with a hole, reducing the number of free carriers. Therefore, it cannot increase the number of electrons.

❓ Why Other Options Are Incorrect

Option A → Correct property of acceptor impurities.
Option B → Correct statement.
Option D → Minority electrons arise through intrinsic generation.

🎯 Strategy

Used

Conceptual Elimination

Application:

Analyze the effect of recombination.

Final Logic:

Recombination reduces carriers, not increases them.

🧠 Memory Trick

"Recombination Removes Carriers."

📚 Class XII – Physics Part II – Chapter: Semiconductor Electronics – Topic: p-Type Semiconductors

14 Correct statements about minority electrons in p-type semiconductors
1. They are intrinsically generated.
2. They are the majority current carriers.
3. Their intrinsic concentration is indirectly reduced by the dopant.
4. They have more chance of meeting majority holes and getting destroyed.

A (A), (C), (D)

B (A), (B), (C)

C (B), (C), (D)

D (A), (B), (D)

💡 Short Explanation

Electrons are minority carriers in p-type material. They arise from intrinsic generation. Many recombine with holes.

✅ Correct Answer Explanation

Electrons in p-type semiconductors are generated intrinsically. Since holes are abundant, electrons have a greater chance of recombining, which reduces their concentration below the intrinsic value.

❓ Why Other Options Are Incorrect

Statement B → Electrons are minority carriers, not majority carriers.

🎯 Strategy

Used

Option Grouping

Application:

Identify correct statements about minority electrons.

Final Logic:

A, C, and D are correct.

🧠 Memory Trick

"p-Type: Holes Plenty, Electrons Few."

📚 Class XII – Physics Part II – Chapter: Semiconductor Electronics – Topic: Majority and Minority Carriers

15 In the energy band diagram of an n-type semiconductor

A each donor atom has an equal charge for certain values of capacitances only

B the donor energy level is located slightly above the top of the valence band

C each donor level allows electrons to move into the conduction band with a very small supply of energy

D each donor state has different charge for any value of its capacitance

💡 Short Explanation

Donor level lies just below the conduction band. Very little energy is required for ionization. Electrons easily enter the conduction band.

✅ Correct Answer Explanation

The donor energy level (Eᴅ) is located very close to the conduction band edge (Eᴄ). Therefore, only a small amount of thermal energy is needed to excite donor electrons into the conduction band.

❓ Why Other Options Are Incorrect

Option A → Irrelevant statement.
Option B → Describes acceptor levels, not donor levels.
Option D → Physically meaningless.

🎯 Strategy

Used

Energy-Level Analysis

Application:

Recall donor-level positioning.

Final Logic:

Donor electrons require very little energy to conduct.

🧠 Memory Trick

"Donor Level Near Conduction Band."

📚 Class XII – Physics Part II – Chapter: Semiconductor Electronics – Topic: Donor Energy Levels

16 Match List I with List II regarding acceptor energy bands

List I	List II
P. Location of Eₐ	1. Gets negatively ionized
Q. Action of valence electron	2. Predominantly due to impurity
R. Ionization of acceptor	3. Jumps to Eₐ with small energy supply
S. Hole density at room temperature	4. Slightly above the top Eᵥ of the valence band

A P-4, Q-3, R-1, S-2

B P-3, Q-4, R-2, S-1

C P-4, Q-1, R-3, S-2

D P-1, Q-2, R-4, S-3

💡 Short Explanation

Acceptor level lies above the valence band. Electrons move into acceptor states. Acceptor ions become negatively charged.

✅ Correct Answer Explanation

Correct matching: P → Slightly above Eᵥ (4) Q → Jumps to Eₐ with small energy supply (3) R → Gets negatively ionized (1) S → Predominantly due to impurity (2)

❓ Why Other Options Are Incorrect

Incorrect assignments of acceptor-level properties.

🎯 Strategy

Used

Option Grouping

Application:

Match acceptor-level concepts.

Final Logic:

Only Option A correctly matches all entries.

🧠 Memory Trick

"EA Above EV, Creates Holes."

📚 Class XII – Physics Part II – Chapter: Semiconductor Electronics – Topic: Acceptor Energy Levels

17 If the intrinsic carrier concentration nᵢ of Si is 1.5 × 10¹⁶ m⁻³ and the number of holes nₕ is 4.5 × 10⁹ m⁻³, what will be the number of electrons nₑ?
\(n_{e}n_{h}=n_{i}^{2}\)

A 5.0 × 10²² m⁻³

B 2.25 × 10³² m⁻³

C 1.5 × 10¹⁶ m⁻³

D 4.5 × 10⁹ m⁻³

💡 Short Explanation

Use the mass-action law: \(n_{e}=\frac{n_{i}^{2}}{n_{h}}\)

✅ Correct Answer Explanation

\(n_{e}=\frac{{\left(1.5\times {10}^{16}\right)}^{2}}{4.5\times {10}^{9}}=\frac{2.25\times {10}^{32}}{4.5\times {10}^{9}}=5.0\times {10}^{22} m^{-3}\)

❓ Why Other Options Are Incorrect

Option B → Equals \(n_{i}^{2}\), not \(n_{e}\).
Option C → Intrinsic concentration only.
Option D → Hole concentration value.

🎯 Strategy

Used

Substitution

Application:

Apply the mass-action law directly.

Final Logic:

\(n_{e}=5.0\times {10}^{22} m^{-3}\)

🧠 Memory Trick

"Mass Action: Product = ni²."

📚 Class XII – Physics Part II – Chapter: Semiconductor Electronics – Topic: Mass Action Law

18 In terms of the energy difference needed to set a fifth electron free versus the intrinsic band gap (Eg) at room temperature, the relation for Si will be

A ~ 0.01 eV >> ~ 1.1 eV

B ~ 1.1 eV << ~ 0.05 eV

C ~ 0.72 eV = ~ 0.72 eV

D ~ 0.05 eV << ~ 1.1 eV

💡 Short Explanation

Donor ionization energy is tiny. Silicon band gap is much larger. Donor electrons ionize easily.

✅ Correct Answer Explanation

The donor electron in Silicon requires approximately 0.05 eV to become free, whereas the intrinsic band-gap energy is about 1.1 eV. \(0.05 eV≪1.1 eV\)

❓ Why Other Options Are Incorrect

Option A → Incorrect comparison.
Option B → Reversed inequality.
Option C → Incorrect values.

🎯 Strategy

Used

Direct Recall

Application:

Compare donor-ionization energy with band-gap energy.

Final Logic:

Donor ionization energy is much smaller than Eg.

🧠 Memory Trick

"0.05 eV vs 1.1 eV."

📚 Class XII – Physics Part II – Chapter: Semiconductor Electronics – Topic: Donor Ionization Energy

19 Incorrect statement about the charge characteristics of an extrinsic semiconductor

A The crystal maintains an overall charge neutrality.

B The charge of additional charge carriers is just equal and opposite to that of the ionized cores.

C A p-type semiconductor is inherently positively charged because holes are the majority carriers.

D Trivalent dopants become effectively negatively charged when sharing the fourth electron.

💡 Short Explanation

Holes are mobile carriers. Overall crystal remains neutral. Majority carriers do not determine net charge.

✅ Correct Answer Explanation

Although holes are the majority carriers in a p-type semiconductor, the crystal remains electrically neutral because ionized acceptor ions balance the charge.

❓ Why Other Options Are Incorrect

Option A → Correct.
Option B → Correct.
Option D → Correct.

🎯 Strategy

Used

Charge Neutrality Principle

Application:

Distinguish carrier type from net charge.

Final Logic:

p-type semiconductors are neutral overall.

🧠 Memory Trick

"Majority Carriers ≠ Net Charge."

📚 Class XII – Physics Part II – Chapter: Semiconductor Electronics – Topic: Charge Neutrality

20 The indirect effect of adding a large number of current carriers of one type

A depends on r

B depends on the angle between r and p

C is to help reduce the intrinsic concentration of minority carriers through increased chance of recombination

D independent of separation of charges

💡 Short Explanation

More majority carriers increase recombination. Minority carrier concentration decreases. This is a key effect of doping.

✅ Correct Answer Explanation

Adding a large number of majority carriers increases the probability that minority carriers will recombine. Consequently, the minority-carrier concentration becomes smaller than its intrinsic value.

❓ Why Other Options Are Incorrect

Option A → Irrelevant quantity.
Option B → Not related to recombination.
Option D → Does not explain minority-carrier reduction.

🎯 Strategy

Used

Cause-and-Effect Analysis

Application:

Relate doping to recombination.

Final Logic:

More majority carriers → more recombination → fewer minority carriers.

🧠 Memory Trick

"More Majority Carriers, Fewer Minority Carriers."

📚 Class XII – Physics Part II – Chapter: Semiconductor Electronics – Topic: Carrier Recombination and Doping Effects

CUET UG Physics Booster Test 2- Extrinsic and Doping Concepts

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